Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS◮Solved 2 ydx − 3 dy2 dx +2y =2e−x , (15.33)subject to the boundary conditions y(0) = 2, y ′ (0) = 1.Taking the Laplace transform of (15.33) and using the table of standard results we obtains 2 ȳ(s) − sy(0) − y ′ (0) − 3 [sȳ(s) − y(0)] +2ȳ(s) = 2s +1 ,which reduces to(s 2 − 3s +2)ȳ(s) − 2s +5= 2s +1 . (15.34)Solving this algebraic equation for ȳ(s), the Laplace transform of the required solution to(15.33), we obtain2s 2 − 3s − 3ȳ(s) =(s +1)(s − 1)(s − 2) = 13(s +1) + 2s − 1 − 13(s − 2) , (15.35)where in the final step we have used partial fractions. Taking the inverse Laplace transformof (15.35), again using table 13.1, we find the specific solution to (15.33) to bey(x) = 1 3 e−x +2e x − 1 3 e2x . ◭Note that if the boundary conditions in a problem are given as symbols, ratherthan just numbers, then the step involving partial fractions can often involvea considerable amount of algebra. The Laplace transform method is also veryconvenient for solving sets of simultaneous linear ODEs with constant coefficients.◮Two electrical circuits, both of negligible resistance, each consist of a coil having selfinductanceL and a capacitor having capacitance C. The mutual inductance of the twocircuits is M. There is no source of e.m.f. in either circuit. Initially the second capacitoris given a charge CV 0 , the first capacitor being uncharged, and at time t =0aswitchinthe second circuit is closed to complete the circuit. Find the subsequent current in the firstcircuit.Subject to the initial conditions q 1 (0) = ˙q 1 (0) = ˙q 2 (0) = 0 and q 2 (0) = CV 0 = V 0 /G, say,we have to solveL¨q 1 + M¨q 2 + Gq 1 =0,M¨q 1 + L¨q 2 + Gq 2 =0.On taking the Laplace transform of the above equations, we obtain(Ls 2 + G)¯q 1 + Ms 2¯q 2 = sMV 0 C,Ms 2¯q 1 +(Ls 2 + G)¯q 2 = sLV 0 C.Eliminating ¯q 2 and rewriting as an equation for ¯q 1 , we findMV 0 s¯q 1 (s) =[(L + M)s 2 + G ][(L − M)s 2 + G ]= V []0 (L + M)s (L − M)s− .2G (L + M)s 2 + G (L − M)s 2 + G502